$A$ convex lens of focal length $0.15 \,m$ is made of a material of refractive index $\frac{3}{2}$. When it is placed in a liquid, its focal length is increased by $0.225 \,m$. The refractive index of the liquid is

$A$ double convex lens has two surfaces of equal radii $15 \, cm$ and refractive index $\mu = 1.5$. Its focal length is equal to ..... $cm$.

$A$ thin lens of focal length $f$ and its aperture has a diameter $d$. It forms an image of intensity $I$. Now, the central part of the aperture up to diameter $(d/2)$ is blocked by an opaque paper. The focal length and image intensity would change to:

$A$ plano-convex lens having a radius of curvature of the first surface $2 \ cm$ exhibits a focal length of $f_1$ in air. Another plano-convex lens with a first surface radius of curvature $3 \ cm$ has a focal length of $f_2$ when it is immersed in a liquid of refractive index $1.2$. If both lenses are made of the same glass of refractive index $1.5$,the ratio of $f_1$ and $f_2$ will be:

$A$ biconvex lens $(R_{1}=R_{2}=20 \ cm)$ has a focal length equal to the focal length of a concave mirror. The radius of curvature of the concave mirror is (Refractive Index of glass lens $= 1.5$) (in $cm$)

Focal length of a convex lens is $20 \ cm$ and its refractive index is $1.5$. It produces an erect,enlarged image if the distance of the object from the lens is: (in $cm$)